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I have just started a course in statistics and have some general questions that have arisen trying to solve the following question:

A survey organisation wants to take a simple random sample in order to estimate the percentage of people who have seen a certain programme. The sample is to be as small as possible. The estimate is specified to be within 1 percentage point of the true value; $\textit{i.e.}$, the width of the interval centered on the sample proportion who watched the programme should be 1%. The population from which the sample is to be taken is very large. Past experience suggests the population percentage to be in the range 20% to 40%. What size sample should be taken?

I think I have to use this and solve for n $$1.96\sqrt{\frac{\pi (1-\pi)}{n}} = .01$$ where $\pi$ is the sample estimated proportion of people who watch the programme.

Now does this mean that I am 95% sure that I am within 1% accuracy? I also am not aware as to how I can find $\pi$ though I have read I could use the population standard deviation instead and suspect I would have to use that as I am given some information- that the pop proportion is 20%-40%.

Finally in general what is being said here:

$$\pi \pm 1.96\sqrt{\frac{\pi (1-\pi)}{n}} = .01$$

My notes at the moment just say it contains the population mean 95% of the time.... why? I think if I had some graphical understanding of what was going on everything would be much simpler for me.

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up vote 1 down vote accepted

Let's get some terminology down here:

  • $\pi$ will denote the true percentage of people that have seen a certain programme. We do not know this value, and are trying to estimate it by sampling people and asking them.
  • $\hat{\pi}_n$ will denote the estimated value that we will get by asking $n$ people and calculating the proportion. This is a random variable.

If we ask one person at random, we know that our response will follow a Bernoulli Distribution $Ber(\pi)$, which we know has a mean of $\mu = \pi$ and a standard deviation of $\sigma = \sqrt{\pi(1-\pi)}$.

The Central Limit Theorem implies that as $n \to \infty$, our estimated proportion $\hat{\pi}_n$ will approach a Normal Distribution with mean $\pi$ and standard deviation $\frac{\sigma}{\sqrt{n}}$.

From here on, we replace "approach" with "is" in the previous sentence, and are basically saying that for a large enough $n$, $\hat{\pi}_n$ is a normal distribution.

Now we can formalize the specific questions you are asking:

What does it mean that I am 95% sure that I am within 1% accuracy? This translates to:

$$ Pr(|\hat{\pi}_n-\pi| \leq 0.01\pi) >= 0.95 $$

$|\hat{\pi}_n-\pi|$ represents how off you are from the truth, and $0.01\pi$ is the most you want to be off. So you want the probability of this happening to be at least 95%.

Now, to solve this we should normalize the left-hand side:

$$ \begin{align} Pr(|\hat{\pi}_n-\pi| \leq 0.01\pi) &= Pr(-0.01\pi \leq \hat{\pi}_n-\pi \leq 0.01\pi) \\ &= Pr\left(\frac{-0.01\pi}{\sigma} \leq \frac{\hat{\pi}_n-\pi}{\sigma} \leq \frac{0.01\pi}{\sigma}\right) \\ &= Pr\left(\frac{-0.01\pi}{\sigma} \leq Z \leq \frac{0.01\pi}{\sigma}\right) \end{align} $$

where $Z$ is the standard normal distribution. We know that this probability is equal to $0.95$ when $$ \frac{0.01\pi}{\sigma} \approx 1.96 $$

So we get to essentially your initial formula (where you interpreted 1% as absolute)

$$ 1.96\sqrt{\frac{\pi(1-\pi)}{n}} = 0.01\pi $$

Since you know that $0.2 \leq \pi \leq 0.4$, you can use the worst-case value of $\pi$, which is 0.2 in this case.

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Assuming that $n$ is large enough to justify a normal approximation to the binomial, and that other assumptions such as independent and unbiased nature of the sample, in frequentist terms, able to say something like:

95% of the times you repeat this exercise, your estimate of the proportion will be within 1 percentage point of the true proportion.

Your second expression $\pi \pm 1.96\sqrt{\dfrac{\pi (1-\pi)}{n}} = .01$ does not really mean anything. If your estimate of the proportion is $\hat\pi$ and you have solved your first expression for $n$ then you might be able to say $$\Pr\left(\hat\pi \in \left(\pi - 1.96\sqrt{\frac{\pi (1-\pi)}{n}} , \pi + 1.96\sqrt{\frac{\pi (1-\pi)}{n}} \right) \right) = 0.95$$ but what you may want to say is the similar but reversed $$\Pr\left(\pi \in \left(\hat\pi - 1.96\sqrt{\frac{\hat\pi (1-\hat\pi)}{n}} , \hat\pi + 1.96\sqrt{\frac{\hat\pi (1-\hat\pi)}{n}} \right) \right) = 0.95.$$

The last of these (which says something like there being a 95% probability of the actual proportion being within one percentage point of the estimated proportion) is not actually meaningful in frequentist terms. Instead you need Bayesian methods.

Since you do not know $\pi$ or before the sample know $\hat\pi$, you might take a worst case for $\pi$ to work out your sample size. This happens when $\pi=\frac12$ in which case you get $n= 1.96^2 \times {\frac14} / 0.01^2 =9604$.

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